https://codeforces.com/contest/1440/problem/C2
The method of C1 is not good enough, and the requirement of C2 is nm. The guess is that each point comes once, and a special judgment is made at the end.
Structure discovery:
First, change all (1,1)~(n-2,m) to 0 each time. Traverse one by one. Then process the last two rows to m-2 column. The last two columns are processed last.
Traverse one by one. For the first case, if the current bit is 1, it will be adjacent to the right and bottom, and if it reaches the boundary, it will be changed to bottom and left.
In the second case, for the row n-1, change itself, right 1, right 2, and for row n, change right 1, right 2. Each time this can make the columns of the current two rows become 0.
Finally, judge the final 2x2, using the function written in C1.
#include<iostream>
#include<vector>
#include<queue>
#include<cstring>
#include<cmath>
#include<map>
#include<set>
#include<cstdio>
#include<algorithm>
#define debug(a) cout<<#a<<"="<<a<<endl;
using namespace std;
const int maxn=120;
typedef int LL;
LL ans=0;
char s[maxn][maxn];
struct P{
LL shux1,shuy1,shux2,shuy2,shux3,shuy3;
};
LL n,m;
vector<P>v;
void mody4(LL x1,LL y1,LL x2,LL y2,LL x3,LL y3,LL x4,LL y4){
v.push_back({x1,y1,x2,y2,x3,y3});
v.push_back({x2,y2,x3,y3,x4,y4});
v.push_back({x1,y1,x2,y2,x4,y4});
v.push_back({x1,y1,x3,y3,x4,y4});
}
void mody3(LL x1,LL y1,LL x2,LL y2,LL x3,LL y3,LL x4,LL y4){
if(s[x1][y1]=='0'){
v.push_back({x2,y2,x3,y3,x4,y4});
}
else if(s[x2][y2]=='0'){
v.push_back({x1,y1,x3,y3,x4,y4});
}
else if(s[x3][y3]=='0'){
v.push_back({x1,y1,x2,y2,x4,y4});
}
else if(s[x4][y4]=='0'){
v.push_back({x1,y1,x2,y2,x3,y3});
}
}
void mody2(LL x1,LL y1,LL x2,LL y2,LL x3,LL y3,LL x4,LL y4){
if(s[x1][y1]=='1'&&s[x2][y2]=='1'){
v.push_back({x2,y2,x3,y3,x4,y4});
v.push_back({x1,y1,x3,y3,x4,y4});
}
else if(s[x3][y3]=='1'&&s[x4][y4]=='1'){
v.push_back({x4,y4,x1,y1,x2,y2});
v.push_back({x3,y3,x1,y1,x2,y2});
}
else if(s[x1][y1]=='1'&&s[x3][y3]=='1'){
v.push_back({x3,y3,x2,y2,x4,y4});
v.push_back({x1,y1,x2,y2,x4,y4});
}
else if(s[x2][y2]=='1'&&s[x4][y4]=='1'){
v.push_back({x4,y4,x1,y1,x3,y3});
v.push_back({x2,y2,x1,y1,x3,y3});
}
else if(s[x1][y1]=='1'&&s[x4][y4]=='1'){
v.push_back({x2,y2,x3,y3,x4,y4});
v.push_back({x1,y1,x2,y2,x3,y3});
}
else if(s[x2][y2]=='1'&&s[x3][y3]=='1'){
v.push_back({x1,y1,x3,y3,x4,y4});
v.push_back({x1,y1,x2,y2,x4,y4});
}
}
void mody1(LL x1,LL y1,LL x2,LL y2,LL x3,LL y3,LL x4,LL y4){
if(s[x1][y1]=='1'){
v.push_back({x1,y1,x2,y2,x3,y3});
v.push_back({x3,y3,x1,y1,x4,y4});
v.push_back({x1,y1,x2,y2,x4,y4});
}
else if(s[x2][y2]=='1'){
v.push_back({x2,y2,x1,y1,x4,y4});
v.push_back({x4,y4,x2,y2,x3,y3});
v.push_back({x2,y2,x1,y1,x3,y3});
}
else if(s[x3][y3]=='1'){
v.push_back({x3,y3,x1,y1,x4,y4});
v.push_back({x4,y4,x2,y2,x3,y3});
v.push_back({x1,y1,x2,y2,x3,y3});
}
else if(s[x4][y4]=='1'){
v.push_back({x4,y4,x2,y2,x3,y3});
v.push_back({x3,y3,x1,y1,x4,y4});
v.push_back({x1,y1,x2,y2,x4,y4});
}
}
void solve(LL x1,LL y1,LL x2,LL y2,LL x3,LL y3,LL x4,LL y4)
{
LL sum1=0;LL sum0=0;
if(s[x1][y1]=='1') sum1++; else sum0++;
if(s[x2][y2]=='1') sum1++; else sum0++;
if(s[x3][y3]=='1') sum1++; else sum0++;
if(s[x4][y4]=='1') sum1++; else sum0++;
if(sum1==4){
mody4(x1,y1,x2,y2,x3,y3,x4,y4);
s[x1][y1]=s[x2][y2]=s[x3][y3]=s[x4][y4]='0';
ans+=4;
}
else if(sum1==3){
mody3(x1,y1,x2,y2,x3,y3,x4,y4);
s[x1][y1]=s[x2][y2]=s[x3][y3]=s[x4][y4]='0';
ans+=1;
}
else if(sum1==2){
mody2(x1,y1,x2,y2,x3,y3,x4,y4);
s[x1][y1]=s[x2][y2]=s[x3][y3]=s[x4][y4]='0';
ans+=2;
}
else if(sum1==1){
mody1(x1,y1,x2,y2,x3,y3,x4,y4);
s[x1][y1]=s[x2][y2]=s[x3][y3]=s[x4][y4]='0';
ans+=3;
}
else if(sum1==0){
s[x1][y1]=s[x2][y2]=s[x3][y3]=s[x4][y4]='0';
ans+=0;
}
}
void presolve()
{
for(LL i=1;i<=n-2;i++){
for(LL j=1;j<=m;j++)
{
if(j<m){
if(s[i][j]=='1'){
s[i][j]=char('0'+(1^(s[i][j]-'0')));s[i+1][j]=char('0'+(1^(s[i+1][j]-'0')));s[i][j+1]=(char)('0'+(1^(s[i][j+1]-'0')));
ans++;
v.push_back({i,j,i+1,j,i,j+1});
}
}
else{
if(s[i][j]=='1'){
s[i][j]=char('0'+(1^(s[i][j]-'0')));s[i+1][j]=char('0'+(1^(s[i+1][j]-'0')));s[i+1][j-1]=char('0'+(1^(s[i+1][j-1]-'0')));
ans++;
v.push_back({i,j,i+1,j,i+1,j-1});
}
}
}
}
for(LL j=1;j<=m-2;j++){
if(s[n-1][j]=='1'){
s[n-1][j]='0';s[n-1][j+1]=char('0'+(1^(s[n-1][j+1]-'0')));s[n][j+1]=char('0'+(1^(s[n][j+1]-'0')));
ans++;
v.push_back({n-1,j,n-1,j+1,n,j+1});
}
if(s[n][j]=='1'){
s[n][j]='0';s[n-1][j+1]=char('0'+(1^(s[n-1][j+1]-'0')));s[n][j+1]=char('0'+(1^(s[n][j+1]-'0')));
ans++;
v.push_back({n,j,n-1,j+1,n,j+1});
}
}
solve(n-1,m-1,n-1,m,n,m-1,n,m);
}
int main(void)
{
cin.tie(0);std::ios::sync_with_stdio(false);
LL t;cin>>t;
while(t--)
{
v.clear();
ans=0;cin>>n>>m;
for(LL i=1;i<=n;i++){
for(LL j=1;j<=m;j++){
cin>>s[i][j];
}
}
presolve();
cout<<ans<<endl;
for(auto i:v){
cout<<i.shux1<<" "<<i.shuy1<<" "<<i.shux2<<" "<<i.shuy2<<" "<<i.shux3<<" "<<i.shuy3<<endl;
}
}
return 0;
}